Three Dimensional Geometry Question 3
Question: A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular from the origin to this plane is given by
Options:
A) $ x^{2}+y^{2}+z^{2}-14=0 $
B) $ x^{2}+y^{2}+z^{2}+x+2y+3z=0 $
C) $ x^{2}+y^{2}+z^{2}-x-2y-3z=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ P(\alpha ,\beta ,\gamma ) $ be the foot of the perpendicular from the origin $ O(0,0,0) $ to the plane so, the plane passes through $ P(\alpha ,\beta ,\gamma ) $ and is perpendicular to OP. clearly direction ratios of OP. i.e,. normal t the plane are $ \alpha ,\beta ,\gamma $ Therefore, equation of the plane is $ \alpha (x-\alpha )+\beta (y-\beta )+\gamma (z-\gamma )=0 $ This plane passes though the fixed point $ (1,2,3) $ , so $ \alpha (1-\alpha )+\beta (2-\beta )+\gamma (3-\gamma )=0 $ or $ {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}-\alpha -2\beta -3\gamma =0 $ Generalizing $ \alpha ,\beta $ and $ \gamma , $ locus of $ P(\alpha ,\beta ,\gamma ) $ is $ x^{2}+y^{2}+z^{2}-x-2y-3z=0 $
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